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Delta attractive potential.

  1. Jul 2, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]V(x) = \begin{cases}
    \infty & , & x<0 \\
    -g*\delta (x-a) & , & x>0
    \end{cases}
    \text{ }[/itex]


    Write the wave function in the left side of x=a and right side of x=a (BOUND STATE)




    3. The attempt at a solution


    In the right side o x=a i would say that it is [itex]\psi (x)=\text{Be}^{-\text{kx}}[/itex].

    But in the left side, i am having problems figuring out what is the wave function.

    Is it [itex]\psi (x)=\text{Be}^{\text{kx}}[/itex] in the even case and [itex]\psi (x)=-\text{Be}^{\text{kx}}[/itex] in the odd case??

    Or because it is between two potential, it is [itex]\psi (x)=A *\text{Cos}[\text{px}][/itex] even case.

    [itex]\psi (x)=A *\text{Sin}[\text{px}][/itex] odd case.
     
    Last edited: Jul 2, 2013
  2. jcsd
  3. Jul 2, 2013 #2

    dextercioby

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    Who's k ?
     
  4. Jul 2, 2013 #3
    k is equal to [itex]\sqrt{ \frac{2 *m |E|}{\hbar^{2}}}[/itex]

    [itex]\frac{d^2\psi }{dx^2}= \frac{2*m*|E|}{\hbar{}^{\wedge}2}\psi[/itex] (BOUND STATE)
     
    Last edited: Jul 2, 2013
  5. Jul 3, 2013 #4
    any help?
     
  6. Jul 3, 2013 #5

    TSny

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    Suppose you start with a simpler case where there is no delt-function potential, so that
    [itex]V(x) = \begin{cases}
    \infty & , & x<0 \\
    0 & , & x>0
    \end{cases}
    \text{ }[/itex]

    What would be the form of the wavefunction for x > 0 that satisfies the boundary condition at x = 0?
     
  7. Jul 4, 2013 #6
    At x>0 Potential is 0

    [itex]\frac{-h{}^{\wedge}2}{2m}\frac{d^2\psi }{dx{}^{\wedge}2} = \text{E$\psi $}[/itex]


    [itex]k= \sqrt{\frac{2m}{h{}^{\wedge}2}E}[/itex]
    bound state, k is real.

    [itex]\psi = \text{Ce}^{-\text{kx}} +\text{De}^{\text{kx}}\text{ }= \text{Ce}^{-\text{kx}} [/itex]

    e^kx blows up in infinity.



    Other possibility is the particle coming from the right without amplitude , be totally reflected in the infinity barrer at x=0, then it goes to plus infinity with amplitude.
     
    Last edited: Jul 4, 2013
  8. Jul 4, 2013 #7

    TSny

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    You can check that ##\small e^{kx}## and ##\small e^{-kx}## do not satisfy the free-particle Schrodinger equation for positive E.

    Shouldn't a free particle have an oscillatory wave function?
     
  9. Jul 4, 2013 #8
    But in the bound states E<0.

    So is satisfied e^kx and e^-kx.

    I thought that only when E>0, the solutions are e^ikx and e^-ikx, i.e, particle has oscillatory motion.
     
  10. Jul 4, 2013 #9

    TSny

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    Yes, You are right. I overlooked that you are asked to find bound states. Without the delta potential there would not be bound states. Sorry about that. So, let's go back to the original problem with the delta-potential in place.

    For 0< x < a, can you find a superposition of ##e^{kx}## and ##e^{-kx}## that will satisfy the boundary condition at x = 0?
     
  11. Jul 4, 2013 #10
    maybe a 2 cosh = e^kx + e^-kx (even case)
     
  12. Jul 4, 2013 #11

    TSny

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    What should ψ(0) equal?
     
  13. Jul 4, 2013 #12
    I guess that should be zero. As in the infinite box.

    My problem is not knowing the "effect" of the infinite potential.

    I know that is left side of delta x<a, we have Psi(x) = Be^kx
     
    Last edited: Jul 4, 2013
  14. Jul 4, 2013 #13

    TSny

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    Both ekx and e-kx satisfy the Schrodinger equation in the region 0<x<a. So, any linear combination of those two functions also satisfies the Schrodinger equation in this region.

    You are correct that the infinite potential at x = 0 implies the boundary condition ψ(0) = 0. What linear combination of ekx and e-kx satisfies this boundary condition?
     
  15. Jul 4, 2013 #14


    well, its ψ(0) = 0 => ekx - e-kx

    it is a hyperbolic sin.
     
  16. Jul 4, 2013 #15

    TSny

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    Yes. And any multiple of sinh(kx) will also satisfy the boundary condition at x = 0 and also satisfy the Schrodinger equation in region I where 0<x<a.

    So, the form of the solution in region I is ψI(x) = Asinh(kx)

    You have already identified the form of the solution for region II (x>a): ψII(x) = Be-kx.

    To relate A to B, we need another boundary condition. What can you say about the wavefunction at x = a?
     
  17. Jul 4, 2013 #16
    ψII(a) = ψI(a)

    ψ'I(a) - ψ'II(a) = -2mg/[itex]\hbar[/itex] * ψI(a)


    Its going to give a transcendent equation.
     
    Last edited: Jul 4, 2013
  18. Jul 4, 2013 #17

    TSny

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    Yes. This equation can be used to relate A and B.

    Would normalization provide another relation between A and B. If so, then A and B are essentially determined at this point.

    I don't quite get the same sign and numerical factor for the right hand side of this equation.
    Yes, this will yield a transcendental equation. Since A and B are already determined, what important information will this equation provide?
     
  19. Jul 4, 2013 #18
    I checked the equation and i get the negative sign in the RHS of the equation. V = -g δ(x-a)
    is going to the right side, and then we multiple it by the -h^2 /2m.


    The transcendental equation is going to provide the values of the energy eigenvalues.

    Sorry my english btw.
     
  20. Jul 4, 2013 #19

    TSny

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    Should the left hand side be
    ψ'I(a) - ψ'II(a)
    or
    ψ'II(a) - ψ'I(a) ?

    Yes. Good.
     
  21. Jul 4, 2013 #20
    You are right.

    I was thinking that the zone II was the left zone.

    Thanks for all the help!
     
    Last edited: Jul 4, 2013
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